question_answer

Unpolarised light of intensity \[32\text{ }W{{m}^{-2}}\] passes through three polarisers such that the transmission axis of last polariser is crossed with the first. If the intensity of the emerging light is \[3\,W{{m}^{-2}},\] then

(i) What is the angle between the transmission axes of the first two polarisers?

(ii) At what angle will the transmitted intensity be maximum?

Or

(i) In what way is diffraction from each slit related to the interference pattern in a double slit experiment?

(ii) Two students are separated by a 7 m partition wall in a room 10 m high. If both light and sound waves can bend around obstacles, how is it that the students are unable to see each other even though they can converse easily?

(iii) When a low flying aircraft passes overhead, we sometimes notice a slight shaking of the picture on our TV screen. Suggest a possible explanation.

           

Answer:

(i) \[\theta =30{}^\circ \] (ii) \[\theta =45{}^\circ C\]             (i) Let us consider 3 polarisers \[{{P}_{1}},\]\[{{P}_{2}}\] and \[{{P}_{3}}.\] Let \[{{I}_{1}}\] be the intensity of polarised light after passing through the first polariser \[{{P}_{1}}\] and \[{{I}_{0}}\] be the intensity of the unpolarised light. Then, \[{{I}_{1}}=\frac{1}{2}{{I}_{0}}\] If \[\theta \] is the angle between the axis of \[{{P}_{1}}\] and \[{{P}_{2}},\] then the intensity of light after passing through \[{{P}_{2}}\] will be\[{{I}_{2}}={{I}_{1}}{{\cos }^{2}}\theta \Rightarrow {{I}_{2}}=\frac{{{I}_{0}}}{2}{{\cos }^{2}}\theta \] (By Malus law) As, \[{{P}_{1}}\] and \[{{P}_{3}}\] are crossed, the angle between axis of \[{{P}_{1}}\] and \[{{P}_{3}}\] will be                         \[\theta '=90{}^\circ -\theta \] \[\therefore \]The intensity of the light emerging from \[{{P}_{3}}\] is \[{{I}_{3}}={{I}_{2}}{{\cos }^{2}}\theta '={{I}_{2}}{{\cos }^{2}}\,(90{}^\circ -\theta )\] \[=\frac{{{I}_{0}}}{2}{{\cos }^{2}}\theta {{\sin }^{2}}\theta =\frac{{{I}_{0}}}{8}{{\sin }^{2}}2\theta \]             \[\therefore \]\[{{\sin }^{2}}2\theta =\frac{8{{I}_{3}}}{{{I}_{0}}}=\frac{8\times 3}{32}=\frac{3}{4}[\because {{I}_{0}}=32\,\,\text{W}{{\text{m}}^{-2}}(\text{given})]\]` Or

(i) \[\sin 2\theta =\sqrt{3}/2\Rightarrow 2\theta ={{\sin }^{-1}}(\sqrt{3}/2)\]                         \[2\theta =60{}^\circ \Rightarrow \theta =30{}^\circ \] \[\therefore \] The angle between the transmission axis of the first two polarisers is \[30{}^\circ .\]

(ii) The intensity of polarised light transmitted from \[{{P}_{3}}\] will be maximum when,             \[\sin 2\theta =\text{maximum=1}\] \[\Rightarrow \]   \[\sin 2\theta =90{}^\circ \Rightarrow 2\theta =90{}^\circ \Rightarrow \theta =45{}^\circ \] \[\therefore \] At angle \[45{}^\circ \] the intensity of the transmitted light will be maximum.

Or

(i) If the width of each slit is comparable to the wavelength of light used, then the interference pattern thus obtained in the double slit experiment is modified by diffraction from each of the two slits.

(ii) As, we know that the frequencies of sound waves lie between 20 Hz to 20 kHz so, their wavelength ranges between 15 m to 15 mm. The diffraction occur, if the wavelength of wave is nearly equal to slit width. As, the wavelength of light waves is \[7000\times {{10}^{-10}}m\] to \[4000\times {{10}^{-10}}m,\]the slit width is very near to the wavelength of sound waves as compared to light waves. Thus, students are unable to see each other even though they can converse easily.

(iii) The weak radar signals sent by the low flying aircraft can interfere with the TV signals received by the antenna. As a result, TV signals may get distorted. Hence, there is slight shaking of the picture on the TV.